First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 2. SOLUTIONS OF EQUATIONS: ROOT-FINDING 80
for g will not be helpful, since g does not satisfy the first condition of Theorem
40: g(x) /∈ [1, 3] for all x ∈ [1, 3] (g(3) = −5 /∈ [1, 3]).
(b) Since p is a root for f, we have p 3 =2p 2 +1,orp =(2p 2 +1) 1/3 . Therefore, p is
the solution to the fixed-point problem g(x) =x where g(x) =(2x 2 +1) 1/3 .
• g is increasing on [1, 3] and g(1) = 1.44,g(3) = 2.67, thusg(x) ∈ [1, 3] for all
x ∈ [1, 3]. Therefore, g satisfies the first condition of Theorem 40.
• g ′ 4x
(x) = and g ′ (1) = 0.64,g ′ (3) = 0.56 and g ′ is decreasing on [1, 3].
3(2x 2 +1) 2/3
Therefore g satisfies the condition in Remark 41 with λ =0.64.
Then, from Theorem 40 and Remark 41, the fixed-point iteration converges if
g(x) =(2x 2 +1) 1/3 .
2. Take λ = k =0.64 in Corollary 42 and use bound (4):
|p − p n |≤(0.64) n max{1 − 1, 3 − 1} =2(0.64 n ).
We want 2(0.64 n ) < 10 −4 , which implies n log 0.64 < −4 log 10 − log 2, or n >
−4 log 10−log 2
log 0.64
≈ 22.19. Therefore n =23is the smallest number of iterations that ensures
an absolute error of 10 −4 .
Julia code for fixed-point iteration
The following code starts with the initial guess p 0 (pzero in the code), computes p 1 = g(p 0 ),
and checks if the stopping criterion |p 1 − p 0 |